Baumslag–Solitar group

Remarks

Small cases: $BS(1, 1) \cong \Z^2$, and $BS(1, -1)$ is the fundamental group of the Klein bottle. The group $BS(1, n)$ is isomorphic to the semidirect product $\Z[1/n] \rtimes \Z$, where the generator of $\Z$ acts by multiplication by $n$. The isomorphism sends $a$ to $1 \in \Z[1/n]$ and $b$ to the generator of $\Z$. In particular $BS(1, n)$ is solvable.

$BS(m, n)$ is residually finite if and only if $\abs{m} = 1$, $\abs{n} = 1$ or $\abs{m} = \abs{n}$.

$BS(2, 3) = \langle a, b \mid b a^2 b^{-1} = a^3 \rangle$ is not Hopfian. Let $\varphi$ be the endomorphism with $\varphi(a) = a^2$ and $\varphi(b) = b$. It is well defined, since $\varphi(b a^2 b^{-1}) = b a^4 b^{-1} = (b a^2 b^{-1})^2 = a^6 = \varphi(a^3)$. It is surjective, since $a = a^3 a^{-2} = b a^2 b^{-1} a^{-2} = \varphi(b a b^{-1} a^{-1})$. It is not injective: the element $[b a b^{-1}, a]$ is not $1$ by Britton's lemma, but $\varphi$ sends it to $[b a^2 b^{-1}, a^2] = [a^3, a^2] = 1$.